# ISI2014-DCG-48

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If $x$ is real, the set of real values of $a$ for which the function $$y=x^2-ax+1-2a^2$$ is always greater than zero is

1. $- \frac{2}{3} < a \leq \frac{2}{3}$
2. $- \frac{2}{3} \leq a < \frac{2}{3}$
3. $- \frac{2}{3} < a < \frac{2}{3}$
4. None of these
in Calculus
recategorized
0
I have solved this by finding Discriminant.

and getting $x = -2a, \;\text {and} \;2a$

Is this the right method?

1 vote

Refer this on how to use discriminant to solve such problems.

For the quadratic function  $ax^{2}+bx+c=0$ to have positive value, the discriminant $b^{2}-4ac<$  $0$

So, on simplifying for the given function, we get   $9$$a^{2}$ $<$ $4$  $\Rightarrow$  $a^{2}$  $<$  $\frac{4}{9}$  $\Rightarrow$  $-\frac{2}{3}$  $<$ $a$  $<$  $\frac{2}{3}$

Option C is the answer.

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